Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 24}{x + 8} = \dfrac{-3x + 16}{x + 8}$
Answer: Multiply both sides by $x + 8$ $ \dfrac{x^2 - 24}{x + 8} (x + 8) = \dfrac{-3x + 16}{x + 8} (x + 8)$ $ x^2 - 24 = -3x + 16$ Subtract $-3x + 16$ from both sides: $ x^2 - 24 - (-3x + 16) = -3x + 16 - (-3x + 16)$ $ x^2 - 24 + 3x - 16 = 0$ $ x^2 - 40 + 3x = 0$ Factor the expression: $ (x - 5)(x + 8) = 0$ Therefore $x = 5$ or $x = -8$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.